I'm having some problems understanding how if a moving object emits light, the speed of that light is still the constant, $c$, speed of light. According to Einstein's theory of relativity, light travels at the same speed for all inertial frames of reference. From what I understand if a driver driving a car at a constant velocity with headlights on sees the light, it would be travelling at $c$, and an observer on the side of the road would also see the light as speed $c$.

At school we have learned about time dilation and length contraction, however, I do not see how they could be applied in this situation to explain why the light "slows down".

I would appreciate it if your answers could try to remain with a knowledge level of a high schooler.

I have also looked at other threads on this exchange, but either I do not understand them or they don't answer the question I have.

  • $\begingroup$ The constancy of light speed is a postulate in SR though it has empirically been proved many times. $\endgroup$ Jul 20, 2020 at 6:24
  • $\begingroup$ It’s just that our intuition about the “addition” of velocities, based on low velocities, is wrong at high velocities. They don’t add. There is a slightly more-complicated formula. $\endgroup$
    – G. Smith
    Jul 20, 2020 at 6:38
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    $\begingroup$ The observable effect of velocity on light is called the Doppler effect; the speed remains c but the frequency changes. en.wikipedia.org/wiki/Relativistic_Doppler_effect $\endgroup$
    – m4r35n357
    Jul 20, 2020 at 9:17
  • $\begingroup$ A ball of light passes by you at a high velocity in your frame of reference. It emits light in all directions at a variety of speeds from your point of view. No matter which direction the light travels across space in your frame of reference, this then results in the speed of light always ending up being 300,000 km/s. Therefore, if the ball of light heads off to your right at 260,000 km/s, it manages to emit light off to the left at 560,000 km/s, which then results in the light moving at 300,000 km/s across space in your frame of reference. But How? Is this similar to your question. $\endgroup$
    – Sean
    Jul 20, 2020 at 20:47
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    $\begingroup$ Time dilation and length contraction are a consequence of the speed of light being the same for all inertial observers. Also, what do you mean by "the light slows down?" $\endgroup$ Jul 20, 2020 at 22:42

5 Answers 5


At school we have learn about time dilation and length contraction however I do not see how they could be applied in this situation to explain why the light "slows down".

This is a little backwards from the usual approach. Usually, we postulate the principle of relativity and the invariance of c and then use that to explain time dilation, length contraction, and the relativity of simultaneity. In principle, it would be possible to instead postulate time dilation, length contraction, and the relativity of simultaneity and use those to derive the usual postulates*, but in practice that is not typically done.

Instead, if you are having trouble with the invariance of the speed of light, you might prefer an approach that does not use it. It turns out that, if you use only the principle of relativity, you can prove that there are two possibilities. Either there is no finite invariant speed, in which case you get Galilean relativity, or there is a finite invariant speed, in which case you get special relativity. Then it becomes a simple matter of experimentally testing to see which of the two possibilities corresponds to reality.

*If you really wanted to derive the invariance of c from time dilation, length contraction, and the relativity of simultaneity, the procedure is straight forward. Time dilation, length contraction, and the relativity of simultaneity together give you the Lorentz transform:

$t’=\gamma t - \gamma v x/c^2$

$x’= \gamma x - \gamma v t$

Then a light pulse in the unprimed frame is given by


so substituting in we get the equation for a light pulse in the unprimed frame

$t’=\gamma t - \gamma v t /c$

$x’=\gamma c t - \gamma v t = c t’$

So the speed of the light pulse is c in the primed frame also. The speed of light is therefore invariant.

  • $\begingroup$ Okay, so assuming from all inertial frames of reference, the speed of light is c, in the example above, how can time dilation, length contraction or mass dilation be used to explain why the speed of light remains constant? Thanks! $\endgroup$
    – Mrocks22
    Jul 21, 2020 at 1:29
  • $\begingroup$ I added a paragraph with the explanation. As I said previously, it is not usually done that way, and I prefer the other approach I mentioned instead. $\endgroup$
    – Dale
    Jul 21, 2020 at 2:10

The second postulate of Special Relativity states that "The speed of light in vacuum, c, is a constant, independent of the relative motion of the source". The phenomenons of length contraction and time dilation are consequences of this postulate, so is the postulate of constancy of $c$ that explains these phenomenons, not the opposite. There are two motivations for this postulate, one experimental and one theoretical.

The experimental evidence for constancy of $c$ comes from the famous Mihchelson- Morley experiment, as you can check here: https://en.wikipedia.org/wiki/Michelson–Morley_experiment.

The theoretical motivation comes from the Maxwell wave equation, which was the motivation used by Einstein to write his first article on relativity published in 1905, whose title is "On the Electrodynamics of Moving Bodies". You can check more about these topics here

Maxwell Wave Equation: https://en.wikipedia.org/wiki/Electromagnetic_wave_equation

Special Relativity Postulates: https://en.wikipedia.org/wiki/Postulates_of_special_relativity


The reason why light can have the same speed in relation to all observers, no mater how those observers are moving relative to each other, is that moving observers do not share a common time frame.

To see this, imagine that your friend throws a bowling ball down an alley at 2 meters per second. After 10 seconds the ball will reach a marker 20 meters along the alley. Now, suppose that when the ball is thrown you walk after it at 1 meter per second. When the ball reaches the marker it is 20m from your friend but only 10m from you, because you have been walking after it. Now suppose that you and your friend disagreed what time it was when the ball reached the marker. Specifically, if to you the ball arrived at the marker after 5 seconds, while to your friend it arrived after 10 seconds, you would both conclude that the speed of the ball was 2 metres per second relative to each of you.

You should be able to see now that the ball can be perceived by each of you as having the same speed if (and only if) you abandon the idea that there is a common time frame for the two of you.

That is how relativity works. if you are moving relative to your friend then the time in your frame all along your direction of travel gets progressively further out of synch with time in your friend's frame.


Yes this is a common confusion among "high schoolers" and not only and initially seems as a paradox from our everyday experience with relative moving objects. The problem goes away if you understand that light is not an object but a wave.

To answer with a paradigm as simply as I can, lets consider the following question and scenario which covers what you actually ask:

In what scenario can you subtract the speeds of a moving light source towards an opposite moving observer towards the light?

According to special relativity c(fixed)=λf=λ/Τ where c is the propagation fixed speed of light in vacuum space, λ the wavelength of light (i.e. distance between two waves) and f the frequency of light (i.e. how many waves per second, color for visible light) with f=1/T where T is the period of time between two light waves, there is no relative speed of light for an observer and the propagation speed of light is absolute, meaning has the same value approximately 3E8 m/s for all observers independent their velocity (i.e. speed and relative direction to a light source).

So No. You cannot add the two velocities and say for example light speed is c=300,000 Km/s and I’m moving towards it with v=20Km/s therefore the relative speed of light is 300,020 Km/s!! NO THAT’S WRONG!!

That what is changing however in the above scenario also deduced from the above formula is, that as you move with a finite speed v towards the light waves you get (T-t) which is equivalent with you seeing shorter wavelength of light thus higher frequency of light. Therefore in the above formula in order to keep c speed always at a fixed value of 3E8m/s the speed of light, if the f=1/(T-t) value goes up because your travelling with a speed v inside the light wave and towards the light source (see figure) then λ value in the equation must go proportionally down so that c speed of light has always the same fixed speed thus 3E8=300,000 Km/s.

So what you would experience would be a Doppler light shift in the frequency of light you see thus a color change. Moving towards the light and you will see the light blue shifted. Moving away from the light and you will see it red shifted.

Nevertheless, there is a loophole on the above arguments which I believe is related to your question and will resolve all your confusion. That in order the above to be true we assume that the light from the light source has already reached you and you can see all this time the light? Thus, you are inside the light waves all this time you are moving. That is the tricky part I believe you are asking and the source of the confusion.

So, is there a scenario where you are allowed to do c-v or c+v for calculating a relative propagation speed of light according to your motion? Yes of course. You are allowed to do that only in the case you have NOT seen yet the light from a light source which was turned on because its light has not reached you yet but it is on its way towards you moving then you are allowed to do any algebraic addition or subtraction depending your relative velocity to calculate the time needed for the light to reach your position in space!! Thus in case you are speeding up towards the coming light is effectively increasing the speed of light you can say to c+v in a vacuum with v being your speed. But if you are already inside the light waves, the speed of light independent of your frame of reference and velocity is always at c=300,000 Km/s. In that case the only thing your motion affects the light is its color thus its frequency. Doppler Shift of Light

  • 1
    $\begingroup$ No, if you are moving at a constant velocity $v$ you will never measure the speed of a light beam as $c+v$ or $c-v$, it will always be $c$. This is the second postulate of Einstein's relativity, and it has been amply confirmed by experiment. $\endgroup$
    – Eric Smith
    May 7, 2021 at 19:38
  • $\begingroup$ Please read more carefully my post and what I'm actually saying. My post agrees with SR and you about the impossibility of c-v and c+v and explains the mechanism of why is that and under what conditions. However, assuming you know the distance between a moving observer towards an initially switched off light source at a time, say the ISS space station and suddenly the ISS turns on a light, since the light has not yet reached the observer you can use c+v to calculate the new distance of the observer from the station at the time the light reaches the observer. $\endgroup$
    – Markoul11
    May 8, 2021 at 9:23
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    $\begingroup$ Whether the light has reached the observer or not, the speed of light in vacuum is $c$, period. This is also true whether you treat the light as waves or particles. $\endgroup$
    – Eric Smith
    May 8, 2021 at 12:36

Banesh Hoffmann, Einstein's collaborator, admits that, originally ("without recourse to contracting lengths, local time, or Lorentz transformations"), the Michelson-Morley experiment directly proved Newton's variable speed of light and disproved the constant speed of light:

"Moreover, if light consists of particles, as Einstein had suggested in his paper submitted just thirteen weeks before this one, the second principle seems absurd: A stone thrown from a speeding train can do far more damage than one thrown from a train at rest; the speed of the particle is not independent of the motion of the object emitting it. And if we take light to consist of particles and assume that these particles obey Newton's laws, they will conform to Newtonian relativity and thus automatically account for the null result of the Michelson-Morley experiment without recourse to contracting lengths, local time, or Lorentz transformations. Yet, as we have seen, Einstein resisted the temptation to account for the null result in terms of particles of light and simple, familiar Newtonian ideas, and introduced as his second postulate something that was more or less obvious when thought of in terms of waves in an ether." Banesh Hoffmann, Relativity and Its Roots, p.92 https://www.amazon.com/Relativity-Its-Roots-Banesh-Hoffmann/dp/0486406768

Even Wikipedia, surprisingly, tells the truth about the Michelson-Morley experiment and the speed of light:

"Emission theory, also called emitter theory or ballistic theory of light, was a competing theory for the special theory of relativity, explaining the results of the Michelson–Morley experiment of 1887. [...] The name most often associated with emission theory is Isaac Newton. In his corpuscular theory Newton visualized light "corpuscles" being thrown off from hot bodies at a nominal speed of c with respect to the emitting object, and obeying the usual laws of Newtonian mechanics, and we then expect light to be moving towards us with a speed that is offset by the speed of the distant emitter (c ± v)." https://en.wikipedia.org/wiki/Emission_theory


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